[Schrödinger]

Hey guys!

"The standard state of a substance at a specified temperature is its pure form at 1 bar". So, the standard state of water at 298 K is liquid water at 298 K and 1 bar. The standard state of Magnesium at 3000 K is probably liquid Mg at 1 bar. Before proceeding any further, please tell me if I've got this right.

Okay, assuming I've understood that correctly :
"...for processes happening at the same temperature, Δ_subH° = Δ_fusH° + Δ_vapH°".

Now how is this possible? The standard state of water at 298 K is liquid water. How can one calculate the fusion or vapourization enthalpies at this temperature AND 1 bar, when, at this temperature the standard state of water is ONLY liquid? In order to have water vapour (or ice for that matter) at 298 K, one needs to adjust the pressure, and this messes with the standard state convention.

This is probably a very silly question. But I'm kinda stumped     

[curiouscat]

"...for processes happening at the same temperature, Δ_subH° = Δ_fusH° + Δ_vapH°".

What's that equation?

[Schrödinger]

Yeah, I guess that's a more context specific doubt. please ignore my previous post. Let me rephrase my doubt.

Consider : H₂O(s) --> H₂O(l)

When one says that the standard enthalpy for this reaction at 298 K is say, 10 kJ/mol, what is meant? As far as my understanding goes, it means that if the reaction is carried out under the constant pressure of 1 bar and the reactants and products are all in standard states corresponding to 298 K (and 1 bar as is the convention), then 10 kJ/mol is the energy consumed. But isnt the standard state of water at 298 K the liquid state? How can we have solid ice at 1 bar pressure and 298 K and call it the standard state?

[curiouscat]

Wikipedia:

"Many standard states are non-physical states, often referred to as "hypothetical states". Nevertheless, their thermodynamic properties are well-defined, usually by an extrapolation from some limiting condition, such as zero pressure or zero concentration, to a specified condition (usually unit concentration or pressure) using an ideal extrapolating function, such as ideal solution or ideal gas behavior, or by empirical measurements."

[Schrödinger]

You usually extrapolate when you know that the graph is heading somewhere and you want to logically locate that point - it is like saying "I will be at home at 5 PM tomorrow, because my weekly routine dictates that". That's extrapolation. But that's kinda hard to do when it comes to phase diagrams isn't it? Lines are totally random.

[Jorriss]

Yeah, I guess that's a more context specific doubt. please ignore my previous post. Let me rephrase my doubt.

Consider : H₂O(s) --> H₂O(l)

When one says that the standard enthalpy for this reaction at 298 K is say, 10 kJ/mol, what is meant? As far as my understanding goes, it means that if the reaction is carried out under the constant pressure of 1 bar and the reactants and products are all in standard states corresponding to 298 K (and 1 bar as is the convention), then 10 kJ/mol is the energy consumed. But isnt the standard state of water at 298 K the liquid state? How can we have solid ice at 1 bar pressure and 298 K and call it the standard state?

It just means the reactants are turned into products under standard conditions. Not that they are changed from their standard states. If I recall, those are the enthalpies of formation.

[Schrödinger]

What I fail to understand is how these standard enthalpies were measured initially when the table of enthalpies wasn't available. The people who found out enthalpies at different temperatures : how did they do it?

[fledarmus]

You measure the standard enthalpies by defining a zero point and measuring any changes from that zero point following state functions.

Defining the zero points are the problem

For enthalpies of formation, the zero points have been defined as the enthalpy of the elements in their normal form at STP. For enthalpies of sublimation, fusion, vaporization, the zero points are defined as the standard state of the compound at STP. If you need to calculate changes in enthalpy between two states, whether by reaction or change of state, or both, you can do it by finding a path between those two points and calculating the change in enthalpy for each point along the path. As long as you are using all state functions (where energy change is independent of path), you can calculate the energy change of an unknown path by following known pathways.

[Schrödinger]

Okay reading your post has kind of opened up a new avenue of thought to me :

Let's say I have CO₂ at 298 K and 1 bar (STP, right? - ofcourse don't bother about the difference between 1 atm and 1 bar). Now I form this from the constituent elements Carbon and Oxygen, which happen to be solid graphite and gas respectively at the same conditions. I carry out the reaction under atmospheric pressure (which again, is the standard pressure) in a bomb calorimeter in an attempt to keep the pressure constant, so that any value I obtain is the standard value at this temperature. But now to measure the heat AT this temperature, I also have to maintain constant T??? How do I do that? Okay, assume I do that too. Now the heat evolved from the ensuing reaction is Δ_rH°₂₉₈, right? This is the ideal way to conduct the experiment to measure standard enthalpy, is it not?

[fledarmus]

Ideally, yes, you would react elemental carbon C(s) and elemental oxygen O₂(g) at 0°C and 1 atmosphere, controlling the temperature at 0°C and measuring the amount of heat you removed to keep the temperature at 0°C. However, as you know, carbon doesn't really react well with oxygen at 0°C, and you would be waiting for several centuries to get a result. Instead, you would measure the energy required to raise the elements to a good reactive temperature, run the reaction at that temperature in a bomb calorimeter so you could measure the amount of heat generated by the reaction, and measure the amount of energy released cooling the products back to room temperature. The total heat of the theoretical process of running the reaction at STP can be calculated by making measurements of the actual processes required to run the reaction at some other temperature and pressure, and adding or subtracting as appropriate the amount of heat involved in changing the temperatures or pressures of the components of the reaction.

Like calculating the distance and direction between your house and the school. There may not be a road leading directly from your house to the school, but you know that if you can get there by going three blocks east, turning left, going two blocks north, turning left, going four blocks west, turning left, going 1 block south, turning left, and going 1 block east, then your school must be one block north of you.

[curiouscat]

Point is, Standard States are a good, consistent and useful way of reporting things.

Dosen't have to actually happen at that state.

[Schrödinger]

Ideally, yes, you would react elemental carbon C(s) and elemental oxygen O₂(g) at 0°C and 1 atmosphere, controlling the temperature at 0°C and measuring the amount of heat you removed to keep the temperature at 0°C. However, as you know, carbon doesn't really react well with oxygen at 0°C, and you would be waiting for several centuries to get a result. Instead, you would measure the energy required to raise the elements to a good reactive temperature, run the reaction at that temperature in a bomb calorimeter so you could measure the amount of heat generated by the reaction, and measure the amount of energy released cooling the products back to room temperature. The total heat of the theoretical process of running the reaction at STP can be calculated by making measurements of the actual processes required to run the reaction at some other temperature and pressure, and adding or subtracting as appropriate the amount of heat involved in changing the temperatures or pressures of the components of the reaction.

Somewhere in the back of my mind that was the exact same idea that I was weaving! Guess I just needed a voice other than my own just to confirm things You can have as many scooby snacks as you like. Thank you

Point is, Standard States are a good, consistent and useful way of reporting things.

Dosen't have to actually happen at that state.

Yeah I know I didn't 'have' to bother myself with this and could have taken it for granted. But it really sucks to accept something without understanding the concept. It would've haunted me forever